To determine the correct description, we analyze the magnetic behavior of each oxygen species based on the number of unpaired electrons in their electronic configurations:

1. $O_{2}$ (Dioxygen): Has 16 electrons. Its molecular orbital configuration ends in $(\pi^{*} 2p_{x}^{1} = \pi^{*} 2p_{y}^{1})$, meaning it has two unpaired electrons and is paramagnetic .
2. $O_{2}^{+}$: Has 15 electrons. It is formed by removing one electron from a $\pi^{*}$ orbital of $O_{2}$, leaving one unpaired electron. Thus, it is paramagnetic.
3. $O_{2}^{-}$ (Superoxide): Has 17 electrons. It has one additional electron in the $\pi^{*}$ orbitals compared to $O_{2}$, leaving one unpaired electron. Thus, it is paramagnetic.
4. $O_{2}^{2-}$ (Peroxide): Has 18 electrons. Both $\pi^{*}$ orbitals are fully occupied $(\pi^{*} 2p_{x}^{2} = \pi^{*} 2p_{y}^{2})$, meaning all electrons are paired. Thus, it is diamagnetic.
5. $O$ (Oxygen atom): Has 8 electrons. Its atomic configuration is $1s^{2} 2s^{2} 2p^{4}$. According to Hund's rule, it has two unpaired electrons in the $2p$ orbitals and is paramagnetic.
6. $O^{+}$ (Oxygen atom cation): Has 7 electrons. Its configuration is $1s^{2} 2s^{2} 2p^{3}$, with three unpaired electrons in the $2p$ orbitals, making it paramagnetic.

Evaluating the options: Option A is incorrect because $O_{2}^{2-}$ is diamagnetic. Option B is incorrect because $O_{2}^{-}$ is paramagnetic. Option C is incorrect because $O_{2}^{2-}$ is diamagnetic. Option D is correct because both $O_{2}^{+}$ and $O_{2}$ are paramagnetic.